Let $\vec{a}$ be a vector in the plane containing vectors $\vec{b}=\hat{i}+2 \hat{j}+\hat{k}$ and $\vec{c}=2 \hat{i}-\hat{j}+\hat{k}$. If $\vec{a}$ is perpendicular to $\hat{i}+\hat{j}+3 \hat{k}$ and its projection on $\vec{b}$ is $3 \sqrt{6}$,then $|\vec{a}|^2=$

Let $\overline{u}, \overline{v}$ and $\overline{w}$ be the vectors such that $|\overline{u}|=1, |\overline{v}|=2, |\overline{w}|=3$. If the projection of $\overline{v}$ along $\overline{u}$ is equal to that of $\overline{w}$ along $\overline{u}$ and $\overline{v}, \overline{w}$ are perpendicular to each other,then $|\overline{u}-\overline{v}+\overline{w}|$ is equal to

$3 \hat{i}-2 \hat{j}-\hat{k}, -2 \hat{i}-\hat{j}+3 \hat{k}$ and $-\hat{i}+3 \hat{j}-2 \hat{k}$ are the position vectors of the vertices $A, B$ and $C$ of a $\triangle ABC$ respectively. If $H$ is its orthocenter,then $\overrightarrow{HA}+\overrightarrow{HB}+\overrightarrow{HC} = $

Let $u = -2 \hat{i} + 2 \hat{j} + \hat{k}$ and $v = \hat{i} - 2 \hat{j} + 2 \hat{k}$. Then the component of $v$ on $u$ is

If $\vec{a}$ and $\vec{b}$ are unit vectors,then the maximum value of $|\vec{a} + \vec{b}| + |\vec{a} - \vec{b}|$ is:

Let $\vec{a}=2\hat{i}+\hat{j}-2\hat{k}$,$\vec{b}=\hat{i}+\hat{j}$ and $\vec{c}=\vec{a}\times\vec{b}$. Let $\vec{d}$ be a vector such that $|\vec{d}-\vec{a}|=\sqrt{11}$,$|\vec{c}\times\vec{d}|=3$ and the angle between $\vec{c}$ and $\vec{d}$ is $\frac{\pi}{4}$. Then $\vec{a}\cdot\vec{d}$ is equal to